Question

Time period of 10 oscillations of a simple pendulum is found to be 20 sec by a stopwatch of resolution 0.25 sec. Length of the string is measured to be 30 cm by a scale of least count 0.1 cm. Percentage error in measurement of acceleration due to gravity g by simple pendulum is? $T=2\pi \sqrt{\frac{l}{g}}$

Answer 1

$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow T^2=4{\pi }^2\frac{l}{g}$
$g=4{\pi }^2\frac{l}{T^2}error=\frac{\Delta g}{g}\times 100\%=(\frac{\Delta l}{l}\times 100+\frac{2\Delta T}{T}\times 100)\%$
$=(\frac{0.1}{30}\times 100+2\times \frac{0.25}{20}\times 100)\%$
$=(0.33+2.5)\%=2.83\%$